# Question #8e5c1

Feb 28, 2016

$+ 0.16 \text{V}$

#### Explanation:

To find ${E}_{c e l l}$ when we do not have standard conditions we use the Nernst Equation:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln Q$

At 298K this simplifies to:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.05916}{n} \log Q$

$n$ is the number of moles of electrons transferred.

$Q$ is the reaction quotient.

To find ${E}_{c e l l}^{\circ}$ and the cell reaction list the ${E}^{\circ}$ values most -ve to most +ve :

$\text{ "E^@"(V)}$

$\stackrel{\textcolor{b l u e}{\leftarrow}}{\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times \times x}}$

$S {n}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s S n \text{ } - 0.136$

$T i {O}^{2 +} + e + 2 {H}^{+} r i g h t \le f t h a r p \infty n s T {i}^{3 +} + {H}_{2} O \text{ } + 0.1$

$\stackrel{\textcolor{red}{\rightarrow}}{\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times \times x}}$

The most +ve 1/2 cell is the one that takes in the electrons so you can see that $T i {O}^{2 +}$ will oxidise $S n$ to $S {n}^{2 +}$

The cell reaction is therefore:

$S {n}_{\left(s\right)} + 2 T i {O}_{\left(a q\right)}^{2 +} + 4 {H}_{\left(a q\right)}^{+} \rightarrow S {n}_{\left(a q\right)}^{2 +} + 2 T {i}_{\left(a q\right)}^{3 +} + 2 {H}_{2} {O}_{\left(l\right)}$

The reaction quotient is therefore given by:

$Q = \frac{\left[S {n}^{2 +}\right] {\left[T {i}^{3 +}\right]}^{2}}{{\left[T i {O}^{2 +}\right]}^{2} {\left[{H}^{+}\right]}^{4}}$

To find ${E}_{c e l l}^{\circ}$ subtract the least +ve 1/2 cell from the most +ve$\Rightarrow$

${E}_{c e l l}^{\circ} = + 0.1 - \left(- 0.136\right) = + 0.236 \text{V}$

Now insert the values into the Nernst Equation:

${E}_{c e l l} = 0.236 - \frac{0.05916}{2} \times \log \left[\frac{\left(0.1\right) {\left(0.1\right)}^{2}}{{\left(0.2\right)}^{2} \left({0.05}^{2}\right)}\right]$

${E}_{c e l l} = 0.236 - \frac{0.05916}{2} \log \left[400\right]$

${E}_{c e l l} = 0.236 - \left[0.02958 \times \left(2.6\right)\right]$

${E}_{c e l l} = 0.236 - 0.077$

${E}_{c e l l} = + 0.16 \text{V}$